Problem: Let $f(x)=-4\text{sin}(x)-9\text{ln}(x)+11$. $f'(x)=$
Answer: Recall that ${\dfrac{d}{dx}[\text{ln}(x)]=\dfrac1x}$ and ${\dfrac{d}{dx}[\text{sin}(x)]=\text{cos}(x)}$. $\begin{aligned} f'(x)&=\dfrac{d}{dx}[-4\text{sin}(x)-9\text{ln}(x)+11] \\\\ &=-4{\dfrac{d}{dx}[\text{sin}(x)]}-9{\dfrac{d}{dx}[\text{ln}(x)]}+\dfrac{d}{dx}[11] \\\\ &=-4\cdot{\text{cos}(x)}-9\cdot{\dfrac1x}+0 \\\\ &=-4\text{cos}(x)-\dfrac9x \end{aligned}$ In conclusion, $f'(x)=-4\text{cos}(x)-\dfrac9x$